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# Higher-Order Derivatives

Suppose you are watching a car race. You can easily determine which car is faster by noticing which one arrives first at the goal. All cars travel the same track, but the faster one does it in the least time. Here you are looking at the change of position versus time. This is what we all know as speed.

Things can move at different speeds, so what about the change of speed versus time? Speed is already a change, so you would be talking about the change of a change! This change of speed versus time is acceleration, and it is also the change of the change of position versus time.

## Meaning of Higher-Order Derivatives

Calculus is all about change. When finding the derivative of a function, you are left with yet another function, which in turn, can be differentiated again under certain circumstances.

The derivatives of functions that have already been differentiated are known as Higher-Order Derivatives.

Take for instance, the derivative of the polynomial function

$f(x) = x^5 +2x^4 -x^2 +4x +1,$

for which you can use the Power Rule to find its derivative, that is

$f'(x) = 5x^4 +8x^3 -2x +4.$

The derivative you just found is another polynomial function, so you can use The Power Rule again to find the derivative of $$f'(x),$$ obtaining

$(f'(x))' = 20x^3+24x^2-2.$

This function you just found is known as the second derivative of $$f.$$ If you keep this notation, you can notice that there are a lot of parentheses. Imagine what will happen if you find the third or the fourth derivative! To address this issue, there are two different notations that can be used when writing Higher-Order Derivatives.

## Rules for Writing Higher-Order Derivatives

More than rules for writing Higher-Order Derivatives, we talk about notation. Having consistent notation is useful so everyone is on the same page when talking about Higher-Order Derivatives.

### Prime Notation for Higher-Order Derivatives

Consider the previous example of a second derivative,

$\left( f'(x)\right)' = 20x^3 +24x^2 -2.$

Rather than using all these parentheses, the primes are placed next to each other, that is

$f''(x) = 20x^3 +24x^2 -2.$

If you differentiate the function once again, you will have a third derivative. This is denoted by using three primes

$f'''(x) = 60x^2 +48x.$

Before proceeding to the fourth derivative, note that if you keep adding primes this would be extremely cumbersome. Rather than adding more primes, a number in placed between parentheses.

$f^{(4)}(x) = 120x +48.$

This notation is kept for further derivatives.

The number is written between parentheses in order to avoid mistaking it for an exponent.

### Leibniz's Notation for Higher-Order Derivatives

Leibniz's notation uses fractional notation rather than primes.

Remember that derivatives are not fractions. They are written like fractions just as a means of notation!

Once again, consider the function of the previous example. Its derivative using Leibniz's notation is written as

$\frac{\mathrm{d}f}{\mathrm{d}x} = 5x^4 +8x^3 -2x +4.$

For the second derivative, you need to place the number 2 in the following way:

$\frac{\mathrm{d}^2 f}{\mathrm{d}x^2} = 20x^3 +24x^2 -2.$

This notation is the same for all the higher-order derivatives, so

$\frac{\mathrm{d}^3f}{\mathrm{d}x^3} = 60x^2 +48x,$

and

$\frac{\mathrm{d}^4f}{\mathrm{d}x^4} = 120x +48$

You can read the notation as follows: The fourth derivative of $$f$$ with respect to $$x$$ four times.

$\frac{\mathrm{d}^4 f}{\mathrm{d}x^4}$

This makes more sense when studying Partial Derivatives, but this is out of the scope of this article!

## Formulas for Higher-Order Derivatives

Most of the time, there is not a formula for finding Higher-Order Derivatives. However, some functions have patterns that can be followed when differentiated.

Consider the following exponential function

$f(x)=e^{2x}.$

You can find its derivative by using the Chain Rule, which gives you

$f'(x)=2e^{2x.}$

For its second derivative you can use the Constant Multiple Rule to obtain

\begin{align} f''(x) &= 2\left( 2e^{2x} \right) \\ &= 2^{2} e^{2x} \\ &= 4e^{2x}.\end{align}

Each further differentiation will multiply the previous derivative by a factor of two, so its fifth derivative for example would be

\begin{align} f^{(5)}(x) &= 2^{5} e^{2x} \\ &= 32e^{2x}. \end{align}

In general, its $$n$$th derivative is given by

$f^{(n)} (x) = 2^n e^{2x}.$

There are more functions whose derivatives relate perfectly to certain patterns, but this is also out of the scope of this article.

## Examples of Higher-Order Derivatives

Finding higher-order derivatives is a straightforward task. Just keep differentiating the function as many times as you need! Here are some examples.

Find the third derivative of

$f(x) = 3x^4 +6x^2 -1.$

You can differentiate this polynomial function using the Power Rule, this will give you

$f'(x) = 12x^3 +12x.$The result is yet another polynomial function, so keep going until you find the third derivative, that is

$f''(x) = 36x^2+12,$

and

$f''(x) = 72x.$

You used The Power Rule three times in a row in the above example! You will not always need to use the same differentiation rule over and over again, here is another example.

Find the third derivative of

$g(x) = \sin{4x}.$

Here you can use the Chain Rule along with the fact that the derivative of the sine function is the cosine function, so

$g'(x) = 4\cos{4x}.$

To find the next derivative, you now need to use the Constant Multiple rule along with the fact that the derivative of the cosine function is the negative sine function, so

\begin{align} g''(x) &= 4 \left( -4\sin{4x}\right) \\ &= -16\sin{4x}.\end{align}

You have the sine function again, so differentiate it again to obtain the third derivative

\begin{align} g'''(x) &= -16 \left( 4\cos{4x} \right) \\ &= -64\cos{4x}. \end{align}

Notice a pattern?

Let's take a look at the derivative of another polynomial function.

Find the fourth derivative of

$h(x) = x^2 +3x +1.$

Since this is a polynomial function you need to apply the Power Rule consecutive times. This will give you

$h'(x) =2x+3,$

then

$h''(x) = 2.$

Note that, at this point, the derivative is now a constant function. Hence, the next derivative (and all further derivatives!) will be equal to 0, that is

$h'''(x) = 0,$

and

$h^{(4)} (x) = 0.$

Higher-order derivatives of polynomial functions will tend to become zero at some point. If you differentiate a polynomial function more times than its degree, the derivative will become zero.

## Applications of higher-order derivatives

### Applications of Higher-Order Derivatives in Mathematics

The second derivative of a function gives us information about the maxima and minima of a function and its concavity. For further information about these topics, you can check out our articles on the following subjects:

### Applications of Higher-Order Derivatives in Science

A prime example of higher-order derivatives is acceleration.

Acceleration is the second derivative of position with respect to time.

This means that if you have a function that describes the position of an object, then its second derivative will describe its acceleration.

The position of an object in time is given by the function

$s(t) = 100-4.9t^2,$

where $$t$$ is measured in seconds and $$s(t)$$ is measured in meters. Find its acceleration.

Here you need to differentiate the position function to find the speed of the object. This can be done with the Power Rule, so

$s'(t) = -9.8t.$

Notice that the units for $$s'(t)$$ are meters per second! That is because you can think of the derivative as a slope, which is rise over run. The rise has units of meters, and the run has units of seconds, so

$\frac{\mbox{rise}}{\mbox{run}} = \frac{\mbox{meters}}{\mbox{second}} = \frac{m}{s}.$

Now, since acceleration is the derivative of the speed, you need to differentiate again to find it, that is

$s''(t) = -9.8.$

But what are the units? Remember that this is the derivative of the speed, which had units of

$$m/s$$. So for acceleration, the units are

$\frac{\mbox{rise}}{\mbox{run}} = \frac{\frac{\mbox{meters}}{\mbox{second}}}{\mbox{second}} = \frac{m}{s^2}.$

The acceleration of the object is the constant $$-9.8 m/s^2$$. This is a free-fall object!

Second-order derivatives are also present in more natural phenomena. These include, but are not limited to:

• The diffusion of heat through a material.
• The propagation of waves.
• The diffusion of a substance in a liquid.
• The mechanics of fluids.

## Higher-Order Derivatives - Key takeaways

• Higher-order derivatives are obtained by differentiating a function more than one time.
• The second derivative is obtained by differentiating two times, the third derivative by differentiating three times, and so on.
• There are two ways of denoting higher-order derivatives.
• The prime notation uses up to three primes. For further derivatives (fourth, fifth, and so on) a number inside parentheses is used instead.
• A second derivative is denoted $$f''(x).$$
• A fifth derivative is denoted $$f^{(5)}(x).$$
• Leibniz's notation uses numbers without parentheses.
• A second derivative is denoted $$\frac{\mathrm{d}^2 f}{\mathrm{d}x^2} .$$
• A fifth derivative is denoted $$\frac{\mathrm{d}^5 f}{\mathrm{d}x^5} .$$
• Higher-order derivatives are used to describe several natural phenomena like the diffusion of heat, the propagation of waves, and the mechanics of fluids.

Higher-order derivatives are functions that are differentiated a repeated amount of times. For example, you obtain a third-order derivative by differentiating a function three times.

To find higher-order derivatives you differentiate a function as many times as needed. For example, if you need a second-order derivative you differentiate the function, then, you differentiate again the result you just obtained.

Generally speaking, there is no formula for higher-order derivatives. Some particular functions can have patterns that can lead to formulas, but these formulas are specific to every function.

Higher-order derivatives are functions that are differentiated a repeated amount of times.  A second derivative, a third derivative, a fifth derivative, and so on, are examples of higher order derivatives.

To find Higher-Order derivatives you need to differentiate a function and keep differentiating each new result until you reach the required order.

## Final Higher-Order Derivatives Quiz

Question

Which of the following denotes the fourth derivative of $$f(x)$$?

$$f^{(4)} (x).$$

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Question

Which of the following denotes the third derivative of $$g(x)$$?

$$g'''(x).$$

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Question

Find the second derivative of $$f(x) = 3x^5.$$

$$f''(x) = 60x^3.$$

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Question

Find the second derivative of $$g(x)= 2x+1.$$

$$g''(x) = 0.$$

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Question

Find the second derivative of $$h(x) = x^{10}.$$

$$h''(x) = 90x^8$$

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Question

Which of the following does not denote the third derivative of $$f(x)$$?

$$f^3 (x).$$

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Question

Which of the following does not denote the fifth derivative of $$f(x)$$?

$$\frac{\mathrm{d}f^5}{\mathrm{d}x^5}.$$

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Question

Find the third derivative of $$f(x) = \sin{x}.$$

$$f'''(x) = -\cos{x}.$$

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Question

Find the fifth derivative of $$g(x) = e^x.$$

$$g^{(5)}(x) = e^x.$$

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Question

Suppose you have a power function $$x^n,$$ where $$n$$ is a positive integer. If you differentiate it ____ times than its degree, its derivative will be 0.

more.

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